An equilateral triangle is inscribed in the parabola y2 = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Asked by Pragya Singh | 1 year ago |  158

##### Solution :-

Let us consider OAB be the equilateral triangle inscribed in parabola y2 = 4ax.

Let AB intersect the x – axis at point C.

Diagrammatic representation of the ellipse is as follows:

Now let OC = k

From the equation of the given parabola, we have,

So, y2 = 4ak

y = ±2$$\sqrt{ak}$$

The coordinates of points A and B are (k, $$2 \sqrt{ak}$$), and (k, $$-2 \sqrt{ak}$$)

AB = CA + CB

$$2 \sqrt{ak}+ 2 \sqrt{ak}$$

$$4\sqrt{ak}$$

Since, OAB is an equilateral triangle, OA2 = AB2.

Then,

k2 + ($$2 \sqrt{ak}$$)2 = ($$4 \sqrt{ak}$$)2

k2 + 4ak = 16ak

k2 = 12ak

k = 12a

Thus, AB = $$4\sqrt{ak}$$

= $$4\sqrt{(a×12a)}$$

$$4\sqrt{12a^2}$$

= $$4\sqrt{(4a×3a)}$$

= 4(2)$$\sqrt{3a}$$

$$8\sqrt{3a}$$

Hence, the side of the equilateral triangle inscribed in parabola

y2 = 4ax is $$8\sqrt{3a}$$.

Answered by Abhisek | 1 year ago

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